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Difference between revisions of "2008 IMO Problems/Problem 3"
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For each sufficiently large prime <math>p</math> of the form <math>4k+1</math>, we shall find a corresponding <math>n</math> satisfying the required condition with the prime number in question being <math>p</math>. Since there exist infinitely many such primes and, for each of them, <math>n \ge \sqrt(p-1)</math>, we will have found infinitely many distinct <math>n</math> satisfying the problem.
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For each sufficiently large prime <math>p</math> of the form <math>4k+1</math>, we shall find a corresponding <math>n</math> satisfying the required condition with the prime number in question being <math>p</math>. Since there exist infinitely many such primes and, for each of them, <math>n \ge \sqrt{p-1}</math>, we will have found infinitely many distinct <math>n</math> satisfying the problem.
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Take a prime <math>p</math> of the form <math>4k+1</math> and consider its "sum-of-two squares" representation <math>p=a^2+b^2</math>, which we know to exist for all such primes. If <math>a=1</math> or <math>b=1</math>, then <math>n=b</math> or <math>n=a</math> is our guy, and <math>p=n^2+1 > 2n+\sqrt(2n)</math> as long as <math>p</math> (and hence <math>n</math>) is large enough.

Revision as of 20:39, 3 September 2008

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For each sufficiently large prime $p$ of the form $4k+1$, we shall find a corresponding $n$ satisfying the required condition with the prime number in question being $p$. Since there exist infinitely many such primes and, for each of them, $n \ge \sqrt{p-1}$, we will have found infinitely many distinct $n$ satisfying the problem.

Take a prime $p$ of the form $4k+1$ and consider its "sum-of-two squares" representation $p=a^2+b^2$, which we know to exist for all such primes. If $a=1$ or $b=1$, then $n=b$ or $n=a$ is our guy, and $p=n^2+1 > 2n+\sqrt(2n)$ as long as $p$ (and hence $n$) is large enough.

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